Calculate the heat of reaction for the following: BrCl (g) --> Br (g) + Cl (g) G

Question

Calculate the heat of reaction for the following:

BrCl(g) --> Br(g) + Cl(g)

Given the following data:

Br2(l) -->Br(g)                              H = +30.9 kJ
                                                                
Br2(g) -->2Br(g)                   H = + 192.9 kJ
                                                                  
Cl2(g) -->2Cl(g)                   H = +243.4 kJ
                                                           
Br2(l) + Cl2(g) -->2BrCl(g)    H = +29.2 kJ

Explanation / Answer

Br2(l) + Cl2(g) -->2BrCl(g)    H = +29.2 kJ so if we have the opposite reaction we have to change the signal ofthe H. 2BrCl(g) --> Br2(l) + Cl2(g)    H = - 29.2 kJ Br2(l) -->Br2(g)                         H = + 192.9 kJ Br2(g) -->2Br(g)                      H = +30.9 kJ Cl2(g) -->2Cl(g)                      H = +243.4 kJ Let's sum everything together... 2BrCl(g) + Br2(l) + Br2(g) +Cl2(g) --> Br2(l) +Cl2(g) + Br2(g)+ 2Br(g)+ 2Cl(g) Now we can cancel everything that is equal in both sides of theequation. Then we'll have: 2BrCl(g) --> 2Br(g) +2Cl(g) And the H is the sum of all H's = (- 29.2 kJ + 192.9 kJ+ 30.9 kJ + 243.4 kJ ) = 438 kJ 2BrCl(g) --> 2Br(g) +2Cl(g)      H = 438kJ Now let's divide the coefficients by 2 and the H as well. BrCl(g) --> Br(g) +Cl(g)       H = 219 kJ Any doubts, just PM me ;)

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